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UY1: RC Circuits

Derive and use charging/discharging equations in RC circuits with clear current sign conventions and time-constant interpretation.

  • University Physics Year 1
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Learning objectives

  • Analyse capacitance, resistance, energy transfer, and transient circuit behaviour.
Why this matters + quick links

This page gives the UY1 working model/result for RC Circuits. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

1) At a glance

  • RC circuits show exponential charging and discharging.
  • Time constant:
τ = RC
  • Charging (from zero):
q(t) = CE(1-e^(-t/RC)), i(t) = (E/R)e^(-t/RC)
  • Discharging:
q(t) = q₀e^(-t/RC), i(t) = -(q₀/RC)e^(-t/RC)
  • Modelling context: these are first-order transients in an ideal lumped circuit. At long times, the capacitor behaves like an open circuit for DC (no current).

Prerequisites: Capacitors And Capacitance, Electromotive Force & Power In Circuits
Next uses: R-L Circuit, L-R-C Series Circuit

2) Setup

Series circuit with resistor R, capacitor C, and source E.

  • Choose current positive in the charging direction.
  • Capacitor voltage: V_C = q/C.
  • KVL for charging: E-iR-q/C = 0.

Assume ideal components and constant R,C.

Common traps (signs + initial/final values)
  • Choose a current direction and stick to it. A negative i(t) during discharge simply means the real current is opposite your chosen positive direction.
  • Check endpoints before doing algebra:
  • Charging: q(0) = 0, i(0) = E/R, and q(∞) = CE.
  • Discharging: q(0) = q₀, i(0) = -q₀/(RC), and q(∞) = 0.
  • Time constant is τ = RC (not L/R).

3) Core derivation/explanation

Using i = dq/dt in charging KVL:

Rdq/dt + q/C = E.

Solving first-order ODE with q(0) = 0:

q(t) = CE(1-e^(-t/RC)),

and

i(t) = dq/dt = (E/R)e^(-t/RC).

So current starts at E/R and decays to 0.

For discharging (source removed):

Rdq/dt + q/C = 0 ⇒ q(t) = q₀e^(-t/RC), i(t) = dq/dt = -(q₀/RC)e^(-t/RC).

Negative sign means actual discharge current is opposite the positive charging direction.

At t = τ, e⁻¹ ≈ 0.368:

  • Charging capacitor has reached 1-1/e ≈ 63.2% of final charge.
  • Discharging capacitor has 36.8% of its initial charge left.

Checks (sanity)

  • Units: RC is seconds.
  • Doubling R or C doubles the response time.

4) Worked example(s)

Given R = 100 kΩ, C = 10 μ F, E = 12 V, initially uncharged.

τ = RC = (1.0 × 10⁵)(1.0 × 10⁻⁵) = 1.0 s.

At t = 2.0 s:

q(t) = CE(1-e⁻²) = (10 × 10⁻⁶)(12)(0.865) = 1.04 × 10⁻⁴ C.

Current:

i(t) = (E/R)e⁻² = (12/(1.0 × 10⁵))(0.135) = 1.62 × 10⁻⁵ A.

5) Practice set (with hints + answers)

  1. In an RC charging circuit, what fraction of final charge is reached at t = τ?
  2. A capacitor discharges with τ = 4.0 s. What fraction of initial charge remains at t = 8.0 s?
  3. If R is doubled and C unchanged, how does τ change?

Hints

  • Use q/CE = 1-e^(-t/τ) for charging.
  • For discharge, use q/q₀ = e^(-t/τ).
  • Time constant scales linearly with both R and C.

Answers

  1. 1-e⁻¹ = 0.632 (63.2%).
  2. e⁻² = 0.135 (13.5%).
  3. τ doubles.

6) Summary + next steps

  • RC dynamics are exponential because capacitor voltage depends on accumulated charge.
  • τ = RC sets the circuit response timescale.
  • Keep a consistent current sign convention, especially during discharge.

Next: Magnetic Field & Motion Of Charged Particles In Magnetic Fields Previous: Electromotive Force & Power In Circuits Back To Electromagnetism